POJ3661——Running

Running
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 5232
 
Accepted: 1938
Description
The cows are trying to become better athletes, so Bessie is running on a track for exactly N (1 ≤ N ≤ 10,000) minutes. During each minute, she can choose to either run or rest for the whole minute.
The ultimate distance Bessie runs, though, depends on her 'exhaustion factor', which starts at 0. When she chooses to run in minute i, she will run exactly a distance of Di (1 ≤ Di ≤ 1,000) and her exhaustion factor will increase by 1 -- but must never be allowed to exceed M (1 ≤ M ≤ 500). If she chooses to rest, her exhaustion factor will decrease by 1 for each minute she rests. She cannot commence running again until her exhaustion factor reaches 0. At that point, she can choose to run or rest.
At the end of the N minute workout, Bessie's exaustion factor must be exactly 0, or she will not have enough energy left for the rest of the day.
Find the maximal distance Bessie can run.
Input
* Line 1: Two space-separated integers: N and M * Lines 2..N+1: Line i+1 contains the single integer: Di
Output
* Line 1: A single integer representing the largest distance Bessie can run while satisfying the conditions. 　
Sample Input

5 2
5
3
4
2
10

Sample Output

9

Source
USACO 2008 January Silver
구간 dp, 나는 이렇게 한다. dp[i]를 설정하면 i분에서 n분까지 최대 도달할 수 있는 거리를 표시한다.
상태 방정식에 대해서 제가 한참 조정을 했어요.
관건은 구간[i,n]에서 시간점 k를 찾아야 한다. k는 그 전에 계속 뛰고 중간에 휴식을 취한 다음에 하위 구간의 dp값을 더해야 한다.

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
  
using namespace std;

const int N = 10010;
int dp[N];
int d[N];
int sum[N];

int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		memset (dp, 0, sizeof(dp));
		sum[0] = 0;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &d[i]);
			sum[i] = sum[i - 1] + d[i];
		}
		for (int i = n - 1; i >= 1; --i)
		{
			for (int k = 0; k <= m; ++k)
			{
				if (k == 0)
				{
					dp[i] = max(dp[i], dp[i + 1]);
					continue;
				}
				if (i + 2 * k - 1 <= n)
				{
					dp[i] = max(dp[i], sum[k + i - 1] - sum[i - 1] + dp[i + 2 * k]);
				}
			}
		//	printf("dp[%d] = %d
", i, dp[i]);
		}
		int ans = 0;
		for (int i = 1; i <= n; ++i)
		{
			ans = max(ans, dp[i]);
		}
		printf("%d
", ans);
	}
	return 0;
}